Question

A screw gauge with a pitch of $$0.5mm$$ and a circular scale with $$50$$ divisions are used to measure the thickness of a thin sheet of aluminum. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is $$0.5mm$$ and the 25th division coincides with the main scale line?
A. $$0.75mm$$
B. $$0.80mm$$
C. $$0.70mm$$
D. $$0.50mm$$

Answer 1

Hint: First calculate the least count by using the equation $$LC={\displaystyle \frac{Pitch}{CircularScaleDivision}}=0.01mm$$ . The negative zero error is $$-0.05mm$$ . Then finally use the equation Reading $$=$$ MSR $$+$$ $$LC$$ $$\times$$ CSR $$\pm$$ Error to reach the solution.

Complete answer:
A device used to determine precisely the diameter of a thin wire or the width of a sheet of metal is the screw gauge. It consists of a U-shaped mount fixed by a screwed pin attached to a thimble. A scale transmitted in mm is inscribed side by side to the axis of the thimble. A screw gauge measures even the tiniest length with exact accuracy with a U shaped metallic mount.
A screw gauge also holds two scales-a key scales and an additional scale-just like Vernier calipers. A millimeter-scale passed to $$0.5mm$$ is the main scale; on the other hand, the extra scale is divided into $$50$$ uniform divisions. The extra scale is on the small metal cap that covers the finger when the screw gauge is sewed and measures the hundredth of the measurement.
Given in the problem,
Pitch $$=0.5mm$$
Circular Scale Division $$=50$$
Main Scale Reading (MSR) $$=0.5mm$$
Circular Scale Reading (CSR) $$=25$$
We know that the equation for the least count is
$$LC={\displaystyle \frac{Pitch}{CircularScaleDivision}}$$
$$LC={\displaystyle \frac{0.5mm}{50}}=0.01mm$$
The negative zero error $$=-5\times \left(0.01\right)mm=-0.05mm$$
The thickness can be calculated by using the equation
Reading $$=$$ MSR $$+$$ $$LC$$ $$\times$$ CSR $$\pm$$ Error
Reading $$=0.5mm+25\times 0.01-\left(-0.05\right)=0.8mm$$

Hence, option B is the correct choice.

Note:
Least count is the smallest value that can be measured by the measuring instrument used (in this case screw gauge). A negative zero error is a negative reading away from the actual reading of $$0.00mm$$ . When performing these experiments it is best to take multiple readings to get more precise results.